Assign oxidation number to the underlined elements in each of the following species:

H4P2O7

Let the oxidation number(O.N.) of P be ‘x’


We know that,


O.N. of H = + 1


O.N. of O = -2


Therefore, we have,


4(+ 1) + 2(x) + 7(-2) = 0


4 + 2x-14 = 0


X = + 5


The O.N. of P is + 5.


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